A2 June 2024 Q2
2.
A uniform rod of length \(28a\) is cut into seven identical rods each of length \(4a\). These rods are joined together to form the rigid framework \(ABCDEA\) shown in Figure 1.
All seven rods lie in the same plane.
The distance of the centre of mass of the framework from \(ED\) is \(d\).
The weight of the framework is \(W\).
The framework is freely pivoted about a horizontal axis through \(C\).
The framework is held in equilibrium in a vertical plane, with \(AC\) vertical and \(A\) below \(C\), by a horizontal force that is applied to the framework at \(A\).
The force acts in the same vertical plane as the framework and has magnitude \(F\).
| Scheme | Marks | AO |
|---|---|---|
| Moments about \(ED\) | M1 | 2.1 |
| e.g. \(4 \times 4a \times 2a\cos 30^\circ + 2 \times 4a \times 4a\cos 30^\circ = 7 \times 4a \times d\) | A1 A1 | 1.1b 1.1b |
| \(32\sqrt{3}a^2 = 28ad \Rightarrow d = \dfrac{8\sqrt{3}}{7}a\) * | A1* | 1.1b |
| (4) |
Notes
M1: Dimensionally correct equation with required terms. Accept use of a parallel axis.
Accept equivalent mass ratio e.g. \(4a\) replaced by 1
A1: Unsimplified equation with at most one error. Allow distances in terms of \(\sin 60^\circ\) or \(\cos 30^\circ\) or equivalent.
N.B. Repeated use of an incorrect distance is only one error.
A1: Correct unsimplified equation. Allow distances in terms of \(\sin 60^\circ\) or \(\cos 30^\circ\) or equivalent
A1*: Obtain given answer from correct working including reference to \(d\).
| Scheme | Marks | AO |
|---|---|---|
| Moments about \(C\) | M1 | 3.1a |
| \(8a \times F = \left(4a\cos 30^\circ - \dfrac{8\sqrt{3}}{7}a\right) \times W\) | A1 | 1.1b |
| \(F = \dfrac{3\sqrt{3}}{28}W\) | A1 | 1.1b |
| (3) | ||
| (7 marks) |
Notes
M1: Dimensionally correct equation with required terms and no extras
Equation should be of the form \(\lambda F = (\mu - d)W\)
A1: Correct unsimplified equation
A1: \(0.19W\) or better (\(0.185576….W\))